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satela [25.4K]
3 years ago
14

How did we see through Venus thick atmosphere?

Physics
2 answers:
RideAnS [48]3 years ago
4 0

Answer:

The Venusian atmosphere supports opaque clouds of sulfuric acid, making optical Earth-based and orbital observation of the surface impossible. Information about the topography has been obtained exclusively by radar imaging. Aside from carbon dioxide, the other main component is nitrogen

snow_lady [41]3 years ago
4 0

Explanation:

Because Venus has such a dense atmosphere which is some 100 times thicker than the earth's. Evidently, Venus is sufficiently close to the Sun that the little carbon dioxide it had in its early, Earth-like atmosphere caused the surface to warm up and leach out more carbon dioxide into the atmosphere.

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If a body is moving with constant linear positive acceleration
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Answer:

D

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It says constant positive linear acceleration, which means that the velocity increases at a constant rate.

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mrs_skeptik [129]
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7 0
3 years ago
A block sliding across a level surface has a mass of 2.5 kg and a mechanical energy of 20 joules. What is its velocity?
tia_tia [17]
Hi!

The energy of the block is 4 m/s

To calculate this, you need to use the equation for kinetic energy. The block is sliding (i.e. it's moving). If the object is sliding across a level surface, the only energy it has is kinetic energy, because there is no change in potential energy (which changes with height). So, the mechanical energy will be pure kinetic energy. The equation is the following, derived from the expression for kinetic energy:

v= \sqrt{ \frac{2*Ke}{m}}=\sqrt{ \frac{2*20 (kg*m^{2}*s^{-2}) }{2,5kg}}=4 m/s

Have a nice day!
8 0
3 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
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